F TESTS AND SUMS OF SQUARES

Transcription

1 F TESTS AND SUMS OF SQUARES Dennis Roberts Some of you will present your class a small end of year unit (as extra material) on simple Analysis of Variance, or ANOVA for short. This is a great idea given A) that it nicely follows from simple two group t tests of differences in means AND B) because of it s popularity in empirical research (especially experiments). As part of this introduction, I have found in the past that a good way to start this ANOVA material is by looking at the concept of SUM OF SQUARES. In ANOVA, variability in an overall data set is partitioned into parts that reflect how much variation there is in the sample MEANS... and how much variation there is WITHIN GROUPS, which is commonly called error or experimental error. I have found that IF students catch on to this notion of sums of squares, that the road thru a beginning unit on ANOVA is a much easier and worthwhile trip. So, this short unit focuses on sums of squares and how that links to ANOVA. SIMPLE t TEST FOR DIFFERENCES IN MEANS The first thing we need to do is to recall the basic setup for a simple t test of the differences in means and to see what the NUMERATOR and the DENOMINATOR of the t test... tell us. Recall that the basic two sample t is: t = (Mean1 - Mean2)/SEdifference in means The numerator reflects how much different the two sample means are... or, in a nutshell... our estimate of how much difference in effect the treatments had. Assuming that the downstairs term stays constant, then the larger is the difference in the sample means, the larger will be our t test value. The larger that t test value is, the greater chance we have to REJECT the null hypothesis. The denominator reflects sampling error. If you look at the formula for the SEdifference in means, a main ingredient in that are the estimates for the variances for EACH of the two groups. In some cases, we assume that there is but ONE common variance for the two populations from which we are sampling... and hence pool the variance estimates into one common term. This term reflects how much variation there is WITHIN the experimental 1

2 groups. Now, given some difference in means in the upstairs term, the greater is the within experimental groups variance, the SMALLER will be your t test value. Thus, having a lot of within group variability is a BAD thing for us... if we want to reject the null hypothesis. So, in a nutshell, the t test could also be shown as: t = Variance in means / Experimental error variance t = How much means vary / How much variance there is within groups So, the hope that we have when we do some analysis of the two sample difference in means case is that our numerator which reflects treatment effects will be relatively large... while the denominator which reflects within group variance or error will be relatively small. The larger is the top part in relation to the bottom part, the larger the t value will be and the better our chances for rejecting the null. SUMS OF SQUARES To examine more carefully the numerator and denominator of a t test, we need to re-examine the basic definition for a variance. In words, the variance is the average of the squared deviations around the mean. The variance can be written as: Variance = (Sum Squared Deviations Around Mean)/ n 1 In a given set of data, as the scores spread out more and more from the mean, the variance becomes a larger and larger quantity. In the variance formula, there is a numerator... and a denominator. What is the numerator? Numerator of Variance = Sum of Squared Deviations around the Mean So, if we have a simple set of data like: 5, 4, 3, 2, and 1... the mean is 3... and the deviations will be 2, 1, 0, -1, and -2 AND the squared deviations will be 4, 1, 0, 1, and 4 and the SUM of the squared deviations will be 10. Thus, for this set of 5 values... our variance calculation would be: Variance = 10 / 4 = is the variance. What is the value of 10? 10 is the NUMERATOR 2

3 OF THE VARIANCE FORMULA... and is called the SUM OF SQUARES. Variance = Sum of Squares / n 1 In short, the sum of squares... or sometimes listed as SS... is simply the numerator of the variance. Nothing more than that. What is the SS or sum of squares for the following data? 10, 8, 6, 4, 2 The mean is The deviations around the mean are: The squared deviations are: The SUM of the squared deviations is: Hence, the SS or sum of squares is: If you answered you are correct! Finally, to know the SS... is to know the NUMERATOR of the variance formula. PARTITIONING SUMS OF SQUARES To start off with, I want to show a very simple experiment type of a layout of data and then show what SS quantities are found and how that is done. Then I will focus more on what happens to these quantities as the data set changes in certain ways. What if we had a very simple 3 group experiment where 15 Ss have been assigned at random... n=5 to each of 3 treatments. Here are some possible data. E1 E2 C

4 Normally, we would show basic descriptive stats on our groups: Variable N Mean StDev E E C What if we just look at ALL of the data, what do we notice? What is seen is that there IS variation in the entire data set... a score of 1 was seen (lowest) in E1... and a score of 7 was seen in E2, which is the highest value. We will name this total variation in a bit. Also notice that there are differences in the means OF the 3 groups... 3 in E1, 5 in E2 and 4 in C. We will later call this variation across the groups. If all the means were the same, that value would be 0. But, it s not 0 in this case. Finally, you see that there is variation WITHIN each group...scores vary in E1 for example, from 5 to 1. In E2, they go from 7 to 3... and in C they go from 6 to 2. We will call this variation WITHIN groups. For this example, there is overall variation, there is variation across (or between) the groups and variation within the groups. Let s be a bit more precise. Total Sum of Squares (SS Tot) The first thing we would calculate is the total sum of squares. Remember, we defined SS as simply the numerator of the variance formula so, what we need to do here is to find the sum of the squared deviations around the mean for the TOTAL set of data. Thus, here we will have all 15 scores as a set... we need to find the mean of that overall set... and then get the squared deviations around the mean of that overall set. BTW, that overall mean is sometimes called the GRAND mean. GRAND only means that it s for all the data together. Sco Dev SqDev

5 Mean = 4 Sum of SqDev = 40 All I did was to put all 15 values into ONE column... find the mean (GRAND mean = 4), subtract 4 from all 15 values to get the deviations around the GRAND mean, then square those deviations and add them up (40). By pooling ALL the data together and find the SS for that overall set of data, we have what is called the total sum of squares or... SS Tot for shot. SS Tot = 40 for this data set. SS Tot is the sum of the squared deviations around the GRAND mean of all data points. Thus, we have the first SS quantity that we need. Now we will look at how we partition or subdivide this SS Tot value into 2 other parts... SS across or between the groups (which reflects how the group means differ) and SS within the groups which represents how much WITHIN variation with see inside of the groups. Let s tackle the SS across or between groups next. Here we need to look at the CELL or group means... and compare them to the GRAND mean. Mean E1 = 3 Mean E2 = 5 Mean C = 4 Grand Mean = 4 How much does the mean of E1 differ from the grand mean? Well, 3-4 = -1. The mean of E1 is 1 point BELOW the grand mean. How much does the mean of E2 differ from the grand mean? 4-4 = 0. In fact, the mean of E2 is the same as the grand mean. Finally, how does the mean of C differ from the grand mean? 5-1 = 1. 5

6 So, E1 has -1 deviation from GM E2 has 0 deviation from GM C has 1 deviation from GM All is fine and good BUT... each group mean is standing the place of the 5 values in that group. That is... the mean of 3 in E1 is representing ALL 5 values in E1. It s like we could substitute 3 for all the values in E1... and if that were the case, then we would have 5 values in E1... that all differ by -1 from the grand mean or GM. Now, SS values are square deviation values... so if we thought about all five values in E1 deviating -1 from the GM, we would need to SQUARE all 5 of these -1 values to have the SS term for the mean of the E1 group FROM the GM. Another way to show this is to simply take the deviation of the E1 mean from the GM... square it BUT then weight it by n for that group. SS between E1 mean and GM = (3-4)^2 times n = -1^2 * 5 = 5 But, what about E2 and C? E2 would look like: SS between E2 mean and GM = (5-4)^2 times 5 = 5 SS between C mean and GM = (4-4)^2 times 5 = 0 Thus, the group means deviate from the GM = 10. SS BG (between groups) will be the squared deviations of the group means from the GM... weighted by the ns in each group. SS BG = 10 for this set of data. We saw that the SS Tot = 40. That is, ignoring variations between the groups or within the groups, the overall SS = 40. Now, we see that PART of that SS Tot can be attributed to the fact that the means of the groups vary (from the grand mean) too. 1/4th of the overall SS Tot... is due to group means varying. Thus, so far we have: SS Tot = SS BG +? 40 = 10 +? Now in this case, it should be obvious what that? value will be... it will be 30 BUT we need to see HOW that 30 comes about. What we have not looked at is how much variation there is within each of the groups. Let s look at the E1 group. 6

7 E1 deve1 SQdevE MTB > sum c8 Sum of SQdevE1 = 10 For E1, the sum of the squared deviations around the mean of E1 = 10. SS for the E1 group is 10. What about for group E2? And C? E2 deve2 SQdevE2 C devc SQdevC For E2, if you add up the squared deviations around the mean of E2, you get 10 for the SS within that group. And, for group C, if you add up the squared deviations around the mean of group C, you get 10 again. So, the SS within group E2 = 10 and the SS within group C = 10. The SS WG is the sum of the squared deviations within EACH group, added together across all groups. The SS WG or within groups = = 30. Let s summarize what we have done. SS Tot = SS BG + SS WG 40 = This SS expression is fundamental in a simple experimental data case... the overall SS can be partitioned into TWO parts, one that reflects how much the group means vary and one that reflects how much variation there is within the groups. This breakdown or partitioning is HOW ANOVA is done and ultimately, how the F test is found and interpreted. Now, recall in the beginning, I reviewed the basic formula for a t test on difference in means of 2 groups. We saw that the numerator of the t test reflected how the means in the two groups differed. Hence the upstairs term in that t test was an estimate of the treatment effect or how the group means differed. We hoped 7

8 of course to find that term to be relatively large... compared to the denominator term which reflected how much variation there was WITHIN the two groups and we called that error or experimental error. The goal in the t test was to find a relatively large numerator... between group means term... compared to a denominator term that was small which reflected sampling error. Differences in means represents treatment effects and variations within groups represents sampling error. We have been working on the partitioning of the total sum of squares into a component that reflects variations across the means of the groups and a component that reflects variations within the groups. It seems logical... if the ANOVA and F tests follow the same general logic and patterns as our simple 2 group t test did, that we will have a numerator term for estimates of treatment effects (SS BG) and a denominator term that will estimate sampling error (SS WG). And, this is exactly what we do... within a experimental data set like the one we have been working with here, we FIRST will find the overall SS Tot and then partition that into a BG part and a WG part. This is where there is this algebraically equality exists: SS Tot = SS BG + SS WG. All simple ANOVA problems will start like this obtaining and partitioning the SS quantities. THEN, from that point, there will be come additional steps we will need to take to transform the SS quantities into an F test or F ratio. The F ratio will be used, just like other t test or chi square tests or z tests... at some point these test statistics become sufficiently large that we reject the null we are using in a particular analysis context. But, before we do that, let s have a closer look at some different data patterns and how these would impact on how the partitioning of the SS Tot quantity. What happens when the means get more and more different? What happens when there is more and more within group variation? If you are able to get the feel for what happens to these SS quantities as conditions (between and within the groups) change, you will be more on your way to better understanding how an ANOVA takes place and HOW you interpret the results. Differences in Means Change Recall the original data situation: 3 groups (E1,E2, C) and n=5 within each group. The data were: Row E1 E2 C

9 Variable N Mean StDev E E C Here we saw that the means were 3, 5, and 4. What if we change the mean in one group... say C... to make the differences in means greater? Let s keep the within group variation the same for each group BUT make the mean for C larger by 3 points... that is... change the mean in C by adding 3 points to each value such that now the mean will be 7 rather than 4. Here s the new data. Row E1 E2 newc Variable N Mean StDev E E newc Now we have means that range from 3 to 7... whereas before we have means that ranged from 3 to 5. Therefore, I have changed the data in terms of means of the groups so that there is a larger spread of the mean values... while keeping the within group variation the same. What do you think should happen in terms of the SS quantities? Well, if I make the mean differences greater while keeping the same within group variation, the SS BG term should get relatively larger while keeping the SS WG term the same. Right? NOTE: To simplify the calculations, I will let Minitab do the calculations... what I want YOU to do is to focus on the impacts of what I have done... don t worry about how Minitab produces the values. The simplest way will be for me to let Minitab actually do the ANOVA calculations... from which we can see what the SS quantities are and see how there are impacted by the changes I have made. For the original data: Source DF SS MS F P Factor Error Total Look at the SS column. We had calculated that the SS Tot = 40, which is what is shown above, and that the partitioning into SS BG and SS WG produced 10 for SS BG and 30 for SS WG. This is what 9

10 Minitab shows. For the data where the C group has a mean +3 larger: Source DF SS MS F P Factor Error Total Now, because the overall data have been forced to vary more, our SS Tot is larger = 70 here. But, what is of importance is that when I make the mean of C larger... but kept the within group variation the same... that larger SS Tot is because there are larger differences amongst the group means. In this case, while the SS WG stays the same, the SS BG increases. In a nutshell, if the means vary more, while keeping within group variability the same, the SS BG gets relatively larger compared to the SS WG or error term. THE LARGER THE MEAN DIFFERENCES GET, ASSUMING THAT WITHIN VARIATION IS CONSTANT, THE GREATER WILL BE THE SS BG COMPARED TO THE SS WG. Variations within Groups Change Let s revisit the original data set again. Row E1 E2 C Variable N Mean StDev E E C Here, the mean differences go from 3 to 5. What I want to do is to create a new 3 group data set... where the mean differences are the same as above, but there is more within group variability. What I do is to add a constant of 10 to each data set (this is totally arbitrary)... so here are the results: Row E1a E2a Ca Variable N Mean StDev 10

11 E1a E2a Ca While all the data points move up by 10, that will not impact on the differences in the means. How can I now change the within group variability? What if I do the following: add 4 points to the top values, 2 points to the next values, leave the means or middle values alone... subtract 2 from the next to lowest values... and subtract 4 from the lowest values. That would make the data sets look like: E1a_1 E2a_1 Ca_ Variable N Mean StDev E1a_ E2a_ Ca_ If you look at the descriptive stats, the differences in the means are the same... but the Sds within each group are larger... each now being 3.16 compared to before. The ranges before were 4 points within each group but now they are 8 points. Clearly, I have spread out the data within each treatment group but kept the mean differences as before. So, what should happen to our SS terms? If the mean differences are the same... then our SS BG should remain the same as originally... but if our within group Sds have gotten larger, then the SS WG should increase. That is, if mean differences are kept constant but within group variation increases, the relative size of the SS WG should be larger compared to the SS BG term. Let s see if that happens. Using the Minitab output: Source DF SS MS F P Factor Error Total We see that the SS BG stays at 10, what it was originally, but the SS WG goes from 30 in the first case to 4 times that or 120 in the current case where I deliberately increased within group variability. Consider a set of data... and the SS BG to be A and the SS WG to be B. 11

12 If we hold within group variation constant... but change the size of the mean differences, we will find: 1. If the mean differences get smaller, the SS BG will get smaller relative to the SS WG. 2. If the mean differences get larger, the SS BG will get larger relative to the SS WG. If we hold the mean differences constant... but change the size of the within group variation, we will find: 3. If the within group variation gets smaller, the SS WG will get smaller relative to the SS BG. 4. If the within group variation gets larger, the SS WG will get larger relative to the SS BG. Question: We have talked before about the SS BG being an estimate of treatment effects and the SS WG being an estimate of sampling error. This is like the t test where upstairs is an estimate of mean differences and downstairs is an estimate of sampling error. To increase our chances of rejecting the null (of no population treatment effects), we would like the SS BG to be relatively larger compared to a relatively small SS WG. Right? So, from the options above... 1 to 4... which one(s) is(are) the best situations if we want to improve our chances of rejecting the null? Those who say 2 AND 3... you are the winners. So, there are two general ways to improve your chances for rejecting the null: First, increase your sample sizes... this would lessen sampling error, or Second, if there were some way to increase the size of the mean differences (making the treatments more potent for example), then this would tend to increase the numerator term that we have been looking at. Of course, in most studies, you have little ability to change how large the differences in means will be but you DO have control over what your sample sizes will be in the groups. So, here as well as in general, improving the chance to reject the null is generally more practically accomplished by increasing ns and therefore reducing sample error. GETTING FROM SS VALUES TO AN F TEST The majority of the above discussion has focused on the concept of sum of squares... total, between groups, and within groups. If you are able to look at a data layout and get some feel for 12

13 how large the group means differ and how much variation there is within the groups, you are on your way to thinking about whether you have much of a chance of rejecting the null... which would state that there are NO differences across the population treatments... that is... mu E1 = mu E2 = mu C. At some point in the overall ANOVA analysis... we will have to make a decision about rejecting or retaining the null. Let s see how that will be done. Look at the following example from above: E1a_1 E2a_1 Ca_ Variable N Mean StDev E1a_ E2a_ Ca_ From out Minitab output, we saw that the SS Tot = 130, SS BG = 10, and therefore the SS WG = 120. Since the SS BG is quite small compared to the SS WG or estimate of error, things are not looking to good at the moment. What we do is to construct an ANOVA summary table. It looks like the following: Source df SS MS Fratio p value BG (treat. effects) WG (samp. error) TOTAL Let s work our way thru this table. df = degrees of freedom. For BG, it is # groups = 3-1 = 2 For WG, we have 5-1 within EACH group... so there are 4 dfs within each group... but we have 3 groups of 4 each. The df for WG in our example will be 12. What about df for the total? Well, there are 15 values all together - 1 = 14 df for the total. So, here you will see that just like the SS terms... the df values are partitioned too. Total df = BG df + WG df 14 = Let s put these in the 13

14 table. Source df SS MS Fratio p value BG (treat. effects) 2 WG (samp. error) 12 TOTAL 14 Next in the table are the SS values. From the above calculations, let s put these in the table too. Source df SS MS Fratio p value BG (treat. effects) 2 10 WG (samp. error) TOTAL df values are like ns... if we think about # of groups, the n - 1 value would be # groups - 1 or 2. For WG, the n for each group is 5, so 5-1 is 4 dfs within EACH group so across 3 groups we have 4*3 = 12 dfs. What about overall? Well, N - 1 = 14 if we were thinking of the overall set of data and deviations around the grand mean (GM). Then we have the SS terms and keep in mind that the SS terms are LIKE the numerator parts of variance formulas. The next column in the table is the MS which equals: Mean Square (or what is the MEAN of the SQUARED deviations... or MS/df or nlike value = MS. So, the MS BG = 10/2 = 5... and the MS WG = 120/12 = 10. NOTE: we don t really do the Total MS value. So, let s fill in the table with what we have so far. Source df SS MS Fratio p value BG (treat. effects) WG (samp. error) TOTAL Now, if we have a MS term that represents a numerator part of a variance formula AND a denominator that is like an n value. So, if we divided a numerator SS value by a denominator that is like n... don t we have a VARIANCE value? Yes... we do. So, the MS 14

15 terms ARE like variances... in fact, they ARE variance estimates. In the case of the MS BG, this is an estimate of the variance of the treatment group means... whereas the MS WG is an estimate of the within group or ERROR variance. This is sounding a lot like the t test we first reviewed where the upstairs term reflects treatment mean differences or variability and the downstairs term reflects out estimate of sampling error. Now, as Michael Buffer (famous boxing ring announcer) would say: LET S GET READY TO RUMBLE... AND CALCULATE THAT F STATISTIC! The F ratio is, er, a ratio of TWO variance estimates: Estimate of Treatment Effect Variance F ratio = Estimate of Sampling Error Variance So, the F ratio is = MS BG / MS WG = 5/10 =.5 Now, unlike t tests that have one df value, F tests (used in ANOVA) have a df value associated with the numerator and a df value associated with the denominator. So, for the ANOVA summary table above, there are 2 dfs for the upstairs term and 12 dfs for the downstairs term. What we say in this case is: We have is an F ratio of.5, where there are 2 and 12 degrees of freedom. So, what does an F distribution with 2 and 12 degrees of freedom look like? Here is what it looks like from Minitab. 15

16 As you can see, the F distribution (this is just one of a huge family) is rather radically + skewed. The question is... how do we interpret the F value that we got... which was.5? Where does it fit within this F distribution? Well,.5 is way over to the LEFT side... close to 0. Since the F ratio is a ratio of 2 variances and variances cannot be negative, the F ratio is always a + value. It could approach a value of 0... the top variance estimate (differences in means) was very very small while the bottom variance (sampling error) was very very large. What we have for out data is a small numerator and a larger denominator of which the F ration is LESS than 1. Now, here s a basic principle when dealing with F ratios in the context of ANOVA. The null hypothesis in this case is that the population means for the E1, E2, and C treatments are all the same. Hence, mue1 = mue2 = muc. What would this F ratio look like IF the null were really true? See below. The expected value for the F ratio when the null is true is 1. Any F ratios less than 1 we can ignore in terms of a hypothesis test in this simple ANOVA case. F ratios that are really small suggest that the differences in the means, under the null model, do not even vary as much from study to study as you might expect by chance alone. Remember, even if the null is true, we will see variations in the mean values from study to study... that s sampling error. So, what we are really looking for in this case... would be an F ratio that is LARGER than 1 (which we don t have of course). In a way, this is like a chi square problem... we look for large chi square values before we are willing to reject the null. Same thing applies here... to reject the null, we will look for a relatively large F ratio... one that is way out along the X scale... to the right. If we were thinking of using an alpha level of what we could do is to find the value along the baseline to the right... where 5% or less of the F ratio values would fall. If we think about starting at the th extreme LEFT... what we need is the 95 percentile rank in this th distribution. We need the 95 percentile rank in an F distribution with 2 and 12 degrees of freedom. Using Minitab, here is that value. F distribution with 2 DF in numerator and 12 DF in denominator P( X <= x ) x To reject the null in this case, we need an F ratio = or > than Clearly, we are NO where near that value for out data so 16

17 we retain the null. If these were real experimental data, we would say that there is not sufficient evidence to reject the null in favor of an alternative that says that not all treatment means are the same. Perhaps we don t like this result but... then are the breaks! Let s look at one additional example. What if we were interested in the differences, if any, amongst 3 different instructional methods for teaching a unit in Apstat. We randomly assign 6 Ss each to each of the 3 treatments. Our data look as follows: Row meth1 meth2 meth Variable N Mean StDev meth meth meth Let s look at the summary data for a moment. We see first, the means from the treatment groups are different. Hence, we know that when we find the SS BG, it will NOT be 0. We also see that there is variation within each treatment group... so we know that the SS WG value will NOT be 0. What we need to do is to partition the SS Tot... into the component parts of SS BG and SS WG. While Minitab will do all of this for us... what would the SS Tot represent? Well, if we found the mean of all 18 values... that would be called the grand mean or GM. If you add up all 18 values and divide by you will obtain: So, we would subtract from each and every one of the 18 values... square that deviation... and then sum up those 18 squared deviation values. To obtain the SS BG, we would need to subtract the GM or grand mean from EACH of the treatment group means... square that deviation... and then multiply by 6 since EACH treatment mean is standing in for and representing 6 values in that treatment group. Finally, to find the SS WG, we would need to go group by group... and subtract each group mean from all the values IN that group... square the deviations... and add. We do this for each of the 3 treatment groups. Here is what Minitab provides us. Source DF SS MS F P Factor Error Total So, we see that: 17

18 SS Tot = SS BG + SS WG = Since there are 3 groups - 1, there are 2 dfs BG. Since there are 6 Ss in each group, we have 6-1 = 5 dfs within EACH group but since there are 3 groups, we have 15 dfs WG. The df Total would be 18-1 = 17. Again, df values are additive too: Tot df = df BG + df WG 17 = Dividing the SS by the appropriate df value produces the MS BG and MS WG terms. Then, to find the F ratio, we divide the MS BG by the MS WG... = /9.21 = F ratio in this study is How will we evaluate it? For this experiment, we will use an F ratio with 2 and 15 degrees of freedom. Like the first example, here is what that theoretical F distribution looks like. 18

19 As you can see, the F distribution with 2 and 15 dfs looks about the same as the one with 2 and 12 dfs. In general, F distributions are radically + skewed. What tends to differ is how far out along the X baseline that + skewed distribution will extend OR what the range of values is between the low and high ends. This F test is a one tail test... we need to isolate the upper 5% if we are using an alpha of.05. The critical value in this case will be: P( X <= x ) x Notice that the critical value here is about the same as it was for the other problem. Since our F ratio is much larger than the CV of we REJECT the null hypothesis of equal treatment means in favor of the alternative that says not all means are the same. But, PLEASE NOTE: Rejecting the null does NOT mean that all treatment means are different. By rejecting the null however, we can be sure that at least ONE is different than the others... while the others could be equal. In order to detect where the specific differences are, we do what are called follow up tests... which is beyond the scope of this handout. But, be assured that software like Minitab and other good packages do these easily. QUICK SUMMARY 1. I first reviewed the notion of a two sample t test and tried to get across the principle that the t value is a ratio of our estimate of mean differences compared to our estimate of sampling error. In one sense, the numerator is the good part whereas the denominator is the bad part. 2. We next looked that the notion of sum of squares. SS is nothing more than the numerator of a formula for the variance. SS have a lower limit of 0 (when all values are the same) to the sky is the limit depending on the actual data. 3. I tried to develop the idea that within a set of experimental data (like 3 groups of 5 Ss each), the SS Tot = SS BG + SS WG. That is, we can look at the overall sum of squares using the entire set of data... and see that we can subdivide or partition that into 2 parts: one part SS BG reflects how the treatment means vary and the other part SS 19

20 WG reflects how much sampling error there is (within group variation). 4. As the means get more and more divergent, the SS BG increases relative to the SS WG. As the variation within the groups gets larger and larger, the SS WG increases relative to the SS BG. 5. SS BG is a numerator concept and SS WG is a denominator concept. While there is not much we can do about the SS BG in terms of impacting or controlling it, we CAN increase n in the downstairs term to impact and control (within limits of resources) sampling error. 6. ANOVA partitions data into the components of BG and WG variation. The summary table turns the SS BG and SS WG terms into variance estimates. 7. The F ratio is a ratio of 2 variance estimates: one variance that estimates the treatment mean differences and the other variance estimates the amount of sampling error. 8. In an experimental data situation, our expectation is that the F ratio will be 1 IF the null of no treatment effects is true. 9. We use an F distribution with df BG and df WG to find a critical value needed to reject the null. The F test is a one tail test (for ANOVA purposes). If our F ratio is = or > than our right end critical value, we reject the null. If it is smaller, we retain. 10. Rejecting the null says that at least ONE of the treatment means is different than the others. They all might be different but not necessarily. SOME FINISHING EXERCISES 1. What do the numerator and denominator of a 2 sample t test tell us? 2. What part of a variance formula is the sum of squares term? 3. What are the min and max possible values for a SS term? 4. How do we get the SS Total for a set of data... say 4 groups and 10 Ss in each group? 5. For #4, how would we obtain the SS BG and SS WG terms? 6. What term is impacted if we change the variation within groups? 20

21 7. What term is impacted if we change how large/small the differences in the group means are? 8. How do we get the degrees of freedom for the BG and WG terms? 9. How are the MS BG and MS WG terms found? 10. How do we get the F ratio? 11. If the F test one tailed or two? 12. What does rejecting the null mean for the F test? Comments or questions about the handout above can be directed to: Dennis Roberts dmr@psu.edu 21